Journal of Algebraic Combinatorics, 20, 243–261, 2004
c 2004 Kluwer Academic Publishers. Manufactured in The Netherlands.
The Bruhat Order on the Involutions
of the Symmetric Group
FEDERICO INCITTI∗
incitti@mat.uniroma1.it
Dipartimento di Matematica, Università “La Sapienza”, P.le Aldo Moro 5, 00185, Rome, Italy
Received December 24, 2002; Revised September 23, 2003; Accepted October 3, 2003
Abstract. In this paper we study the partially ordered set of the involutions of the symmetric group Sn with
the order induced by the Bruhat order of Sn . We prove that this is a graded poset, with rank function given by
the average of the number of inversions and the number of excedances, and that it is lexicographically shellable,
hence Cohen-Macaulay, and Eulerian.
Keywords: symmetric group, Bruhat order, involution, E L-shellability, Cohen-Macaulay
1.
Introduction
It is well-known that the symmetric group Sn ordered by the Bruhat order encodes the
cell decomposition of Schubert varieties (see, e.g., [8]). This partially ordered set has been
studied extensively (see, e.g., [4, 5, 7, 11, 12, 19]) and it is known that it is a graded
poset, with rank function given by the number of inversions, and that it is lexicographically
shellable, hence Cohen-Macaulay, and Eulerian.
In [14] and [15] a vast generalization of this partial order has been considered, in relation
to the cell decomposition of certain symmetric varieties. In this work we study this partial
order in a special case that is particularly attractive from a combinatorial point of view,
namely that of the involutions of Sn with the order induced by the Bruhat order (see [14,
Section 10]). Our main results are that this is a graded poset, with rank function given
by the average of the number of inversions and the number of excedances, and that it is
lexicographically shellable and Eulerian.
The organization of the paper is as follows. In Section 2 we give basic definitions, notation
and results that will be needed later. Sections 3 and 4 are devoted to the study of some
new combinatorial concepts, namely those of “suitable rise”, “covering transformation”
and “minimal covering transformation” of an involution, that play a crucial role in the
sequel. In Sections 5, 6 and 7 we prove our main results, namely that the poset is graded,
lexicographically shellable and Eulerian.
∗ This work is part of author’s Ph.D. dissertation, written at “La Sapienza” University of Rome, under the direction
of Professor Francesco Brenti.
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Preliminaries
In this section we recall some basic definitions, notation and results that will be used in the
rest of this work.
We let N = {1, 2, 3, . . .} and Z be the set of integers. For every n ∈ N we let [n] =
{1, 2, . . . , n} and for every n, m ∈ Z, with n ≤ m, we let [n, m] = {n, n + 1, . . . , m}.
2.1.
Posets
We follow [16, Chap. 3] for poset notation and terminology. In particular we denote by the covering relation: x y means that x < y and there is no z such that x < z < y.
The Hasse diagram of a finite poset P is the graph whose vertices are the elements of P,
whose edges are the covering relations, and such that if x < y, then y is drawn “above”
x. A poset is said to be bounded if it has a minimum and a maximum, denoted by 0̂ and 1̂
respectively. For a bounded poset P we denote by P̄ the subposet P\{0̂, 1̂}. If x, y ∈ P,
with x ≤ y, we let [x, y] = {z ∈ P : x ≤ z ≤ y}, and we call it an interval of P. If
x, y ∈ P, with x < y, a chain from x to y of length k is a (k + 1)-tuple (x0 , x1 , . . . , xk )
such that x = x0 < x1 < · · · < xk = y, denoted simply by “x0 < x1 < · · · < xk ”. A chain
x0 < x1 < · · · < xk is said to be saturated if all the relations in it are covering relations,
and in this case we denote it simply by “x0 x1 · · · xk ”.
A poset is said to be graded of rank n if it is finite, bounded and if all maximal chains
of P have the same length n. If P is a graded poset of rank n, then there is a unique rank
function ρ : P → [0, n] such that ρ(0̂) = 0, ρ(1̂) = n and ρ(y) = ρ(x) + 1 whenever y
covers x in P. Conversely, if P is finite and bounded, and if such a function exists, then P
is graded of rank n.
If P is a graded poset and Q is a totally ordered set, an edge-labelling of P with values
in Q is a function λ : {(x, y) ∈ P 2 : x y} → Q. If λ is an edge-labelling of P, for every
saturated chain x0 x1 · · · xk we set
λ(x0 , x1 , . . . , xk ) = (λ(x0 , x1 ), λ(x1 , x2 ), . . . , λ(xk−1 , xk )).
An edge-labelling λ of P is said to be an E L-labelling if for every x, y ∈ P, with x < y,
the following properties hold:
(1) there is exactly one saturated chain from x to y, say x = x0 x1 · · · xk = y, such
that λ(x0 , x1 , . . . , xk ) is a non-decreasing sequence (i.e., λ(x0 , x1 ) ≤ λ(x1 , x2 ) ≤ · · · ≤
λ(xk−1 , xk ));
(2) any other saturated chain from x to y, say x = y0 y1 · · · yk = y, different from the
previous one, is such that λ(x0 , x1 , . . . , xk ) < L λ(y0 , y1 , . . . , yk ), where < L denotes
the lexicographic order ((a1 , a2 , . . . , ak ) < L (b1 , b2 , . . . , bk ) if and only if ai < bi ,
where i = min{ j ∈ [k] : a j = b j }).
A graded poset P is said to be lexicographically shellable, or E L-shellable, if it has an
E L-labelling.
BRUHAT ORDER OF THE INVOLUTIONS OF Sn
245
Connections between E L-shellable posets and shellable complexes, Cohen-Macaulay
complexes and Cohen-Macaulay rings can be found, for example, in [1–3, 9, 10, 13, 17].
Here we only recall some basic facts. The order complex (P) of a poset P is the simplicial
complex of all chains of P. A poset P is said to be shellable if (P) is shellable, and
Cohen-Macaulay if (P) is Cohen-Macaulay (see, e.g., [3, Appendix], for the definitions
of a shellable complex and of a Cohen-Macaulay complex). Furthermore, a poset is CohenMacaulay if and only if the Stanley-Reisner ring associated with it is Cohen-Macaulay (see,
e.g., [13]). It is known that if a complex is shellable, then it is Cohen-Macaulay (see [9,
Remark 5.3]). So the same holds for posets. Finally, Björner has proved the following (see
[3, Theorem 2.3]).
Theorem 2.1 Let P be a graded poset. If P is E L-shellable then P is shellable and
hence Cohen-Macaulay.
A graded poset P with rank function ρ is said to be Eulerian if
|{z ∈ [x, y] : ρ(z) is even}| = |{z ∈ [x, y] : ρ(z) is odd}|,
for every x, y ∈ P such that x < y.
In an EL-shellable poset there is a necessary and sufficient condition for the poset to be
Eulerian. We state it in the following form (see [3, Theorem 2.7] and [18, Theorem 1.2] for
proofs of more general results).
Theorem 2.2 Let P be a graded E L-shellable poset and let λ be an E L-labelling of P.
Then P is Eulerian if and only if for every x, y ∈ P such that x < y, there is exactly one
saturated chain from x to y with decreasing labels.
Finally, we refer to [10] or [17] for the definition of a Gorenstein poset, just recalling the
following result (see, e.g., [17, Section 8]).
Theorem 2.3 Let P be a graded Cohen-Macaulay poset. Then P is Gorenstein if and
only if the subposet of P induced by P\{x ∈ P̄ : x is comparable with every y ∈ P} is
Eulerian.
2.2.
The Bruhat order on the symmetric group
Given a set T we let S(T ) be the set of all bijections π : T → T . As ususal we denote by
Sn = S([n]) the symmetric group, and we call its elements permutations.
If σ ∈ Sn then we write σ = σ1 σ2 · · · σn , to mean that σ (i) = σi for every i ∈ [n]. We
also write σ in disjoint cycle form (see, e.g., [16, p. 17]), omitting to write the 1-cycles of
σ . For example, if σ = 364152, then we also write σ = (1, 3, 4)(2, 6). Given σ, τ ∈ Sn ,
we let σ τ = σ ◦ τ (composition of functions) so that, for example, (1, 2)(2, 3) = (1, 2, 3).
Given σ ∈ Sn , the diagram of σ is a square of n × n cells, with the cell (i, j) (i.e., the
cell in column i and row j, with the convention that the first column is the leftmost one and
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the first row is the lowest one) filled with a dot if and only if σ (i) = j. The diagonal of the
diagram is the set of cells {(i, i) : i ∈ [n]}.
Let σ ∈ Sn . As usual we denote by
Inv(σ ) = {(i, j) ∈ [n]2 : i < j, σ (i) > σ ( j)}
the set of inversions of σ and by inv (σ ) their number.
The (strong) Bruhat order of Sn is the partial order relation on Sn , denoted by ≤ B , which
is the transitive closure of the relation → defined by
σ → τ ⇔ there is a transposition (i, j) such that τ = σ (i, j)
and inv (σ ) ≤ inv(τ ).
Let σ ∈ Sn . A rise of σ is a pair (i, j) ∈ [n]2 such that i < j and σ (i) < σ ( j). A rise (i, j)
of σ is said to be free if there is no k ∈ [n] such that i < k < j and σ (i) < σ (k) < σ ( j).
The following is a well-known result.
Proposition 2.4 Let σ, τ ∈ Sn . Then σ τ in the Bruhat order of Sn if and only if
τ = σ (i, j), for some free rise (i, j) of σ .
Let σ ∈ Sn . For every (h, k) ∈ [n]2 we set
σ [h, k] = |{i ∈ [h] : σ (i) ∈ [k, n]}|.
A fundamental characterization of the Bruhat order relation in Sn is the following (see,
e.g., [11]).
Proposition 2.5
Let σ, τ ∈ Sn . Then σ ≤ B τ if and only if
σ [h, k] ≤ τ [h, k],
for every (h, k) ∈ [n]2 .
A consequence of Proposition 2.5 is the following.
Proposition 2.6
Let σ, τ ∈ Sn . If σ ≤ B τ then σ −1 ≤ B τ −1 .
We are interested in the set of involutions of Sn , which we denote by Invol(n). Note
that a permutation is an involution if and only if its diagram is symmetric with respect to
the diagonal. We wish to study the poset (Invol(n), ≤ B ) of the involutions with the order
induced by the Bruhat order of Sn . We will simply denote by Sn and Invol(n) the respective
posets with the Bruhat order.
In figures 1 and 2 are represented, respectively, the Hasse diagram of the poset S4 , with
the involutions marked, and the Hasse diagram of the poset Invol(4).
BRUHAT ORDER OF THE INVOLUTIONS OF Sn
Figure 1.
Hasse diagram of S4 .
Figure 2.
Hasse diagram of Invol(4).
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It is well-known that Sn is a graded poset, with rank function given by the number of
inversions, and that it is Eulerian (see, e.g., [19, p. 395]). It has been also shown that Sn
is E L-shellable (see, e.g., [7]). Our goal is to prove that similar results hold for the poset
Invol(n).
It should be mentioned that the set of fixed-point-free involutions of Sn , partially ordered
by the Bruhat order, has been recently studied in [6].
3.
Suitable rises and covering transformation
In this section we introduce the concepts of “suitable rise” and “covering transformation”
of an involution, which play a crucial role in the description of the covering relation in
Invol(n).
Let σ ∈ Sn . We denote by
I f (σ ) = Fix(σ ) = {i ∈ [n] : σ (i) = i},
Ie (σ ) = Exc(σ ) = {i ∈ [n] : σ (i) > i},
Id (σ ) = Def (σ ) = {i ∈ [n] : σ (i) < i},
respectively the sets of fixed points, of excedances and of deficiencies of σ . As usual we
denote by exc(σ ) the number of excedances of σ .
The type of a rise (i, j) is the pair (a, b), where a, b ∈ { f, e, d} are such that i ∈ Ia (σ ) and
j ∈ Ib (σ ). A rise of type (a, b) is also called an ab-rise. Furthermore, we need to distinguish
between two kinds of ee-rises: an ee-rise (i, j) is crossing if i < σ (i) < j < σ ( j), noncrossing if i < j < σ (i) < σ ( j). In other words, an ee-rise (i, j) is crossing if the cells
(i, σ ( j)) and ( j, σ (i)) are on opposite sides of the diagonal, and non-crossing otherwise.
For example, if σ = 321654, the free rises of σ are (1, 4), (1, 5), (1, 6), (2, 4), (2, 5),
(2, 6), (3, 4), (3, 5), (3, 6), whose types are, respectively, (e, e), (e, f ), (e, d), ( f, e), ( f, f ),
( f, d), (d, e), (d, f ), (d, d) (all nine possible types) and the ee-rise is crossing.
If σ ∈ Invol(n), then with every ee-rise of σ is associated a symmetric dd-rise, with
every e f -rise a d f -rise, with every f e-rise an f d-rise and with every ed-rise a de-rise. This
justifies the following definition.
Definition 3.1 Let σ ∈ Invol(n). A rise (i, j) of σ is suitable if it is free and if its type is
one of the following: ( f, f ), ( f, e), (e, f ), (e, e), (e, d).
We now introduce the main concept of this section.
Definition 3.2 Let σ ∈ Invol(n). Let (i, j) be a suitable rise of σ . Depending on the type of
(i, j), we define a new involution obtained from σ , which we call covering transformation
of σ with respect to (i, j), and denote by ct(i, j) (σ ), as described in Table 1.
The fourth column in Table 1 (with caption “Move”) describes the action of ct(i, j) on
the diagram of σ : the black dots mark the positions of the elements of σ which move in
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BRUHAT ORDER OF THE INVOLUTIONS OF Sn
Table 1.
Covering transformation.
Case
Type of (i, j)
ct(i, j) (σ )
1
f f -rise
σ (i, j)
2
f e-rise
σ (i, j, σ ( j))
3
e f -rise
σ (i, j, σ (i))
4
Non-crossing ee-rise
σ (i, j)(σ (i), σ ( j))
5
Crossing ee-rise
σ (i, j, σ ( j), σ (i))
6
ed-rise
σ (i, j)(σ (i), σ ( j))
Move
the transformation, while the white dots mark the positions of the new elements after the
transformation. Inside the grey areas there are no other dots of σ . Note that in every case
ct(i, j) (σ ) (i) = σ ( j) .
Lemma 3.3 Let σ, τ ∈ Invol(n), with σ ≤ B τ . Let (i, j) be a suitable rise of σ, but not
an ed-rise, such that σ (i, j) ≤ B τ . Then
ct(i, j) (σ ) ≤ B τ.
Proof: In case 1, ct(i, j) (σ ) = σ (i, j), and there is nothing to prove.
In all other cases (2, 3, 4 and 5), we let σ1 = σ (i, j), σ2 = σ1−1 = σ (σ (i), σ ( j)) and
σ3 = ct(i, j) (σ ). We have σ1 ≤ B τ and, by Proposition 2.6, σ2 ≤ B τ . We want to show that
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σ3 ≤ B τ . In each of these four cases it’s easy to check that for every (h, k) ∈ [n]2 we have
σ3 [h, k] =
σ1 [h, k],
σ2 [h, k],
if h > k,
if h ≤ k.
So, in every case, σ3 [h, k] ≤ τ [h, k] and this implies σ3 ≤ B τ .
It remains to consider case 6 (ed-rise). Note that Lemma 3.3 does not hold in general if
(i, j) is a free ed-rise. For example, if we consider σ = 351624, τ = 653421 and the free edrise (1, 6) of σ , we have σ ≤ B τ and σ (1, 6) = 451623 ≤ B τ , but ct(1,6) (σ ) = 456123 B τ .
We need to introduce a further new concept to handle ed-rises, and this is what we do in
the next section.
4.
Minimal covering transformation
In this section we introduce the concept of “minimal covering transformation” of an involution, crucial in the proofs that the poset Invol(n) is graded and E L-shellable.
Definition 4.1 Let σ, τ ∈ Invol(n), with σ B τ . The difference index of σ with respect
to τ , denoted by di τ (σ ) (or simply di), is the minimal index on which σ and τ differ:
di τ (σ ) = min{i ∈ [n] : σ (i) = τ (i)}.
Note that
di ≤ σ (di) < τ (di).
To prove the two inequalities, first note that, by definition, σ (di) = τ (di). Since σ and
τ are involutions, they must differ at the index σ (di), and di is the minimal index at which
they differ. Thus di ≤ σ (di). To prove the second inequality, suppose, to the contrary, that
σ (di) > τ (di). In this case we would have σ [di, σ (di)] = τ [di, σ (di)] + 1. But σ B τ
and, by Proposition 2.5, this implies σ [di, σ (di)] ≤ τ [di, σ (di)], which is a contradiction.
Thus σ (di) < τ (di).
Definition 4.2 Let σ, τ ∈ Invol(n), with σ B τ . The covering index of σ with respect to
τ , denoted by ci τ (σ ) (or simply ci), is
ci τ (σ ) = min{ j ∈ [di + 1, n] : σ ( j) ∈ [σ (di) + 1, τ (di)]}.
To show that ci is well-defined, set k = σ (τ (di)). If k ∈ [di − 1], then σ (k) = τ (k), or
equivalently k = di, which is a contradiction. If k = di, then σ and τ agree at the index
di, which is also a contradiction. Thus k ∈ [di + 1, n]. Also, σ (k) = τ (di), which implies
that ci is the minimum of a nonempty set.
By definition, (di, ci) is a free rise of σ , but more is true.
BRUHAT ORDER OF THE INVOLUTIONS OF Sn
251
Proposition 4.3 Let σ, τ ∈ Invol(n), with σ B τ . Then (di, ci) is a suitable rise of σ
and
σ (di, ci) ≤ B τ.
Proof: We may assume, without loss of generality, that di = 1. As we have already
observed, di is necessarily either a fixed point or an excedance of σ . So, for the first part,
we only have to show that (1, ci) is not an f d-rise. Suppose, to the contrary, that (1, ci) is
an f d-rise, so σ (1) = 1 and σ (ci) < ci. By definition of ci, there is no k ∈ [2, ci − 1]
such that σ (k) ∈ [1, τ (1)]. In particular, since σ (ci) ∈ [2, ci − 1], we have ci > τ (1). Thus
|{k ∈ [τ (1)] : σ (k) ∈ [τ (1)]}| = 1, which implies
σ [τ (1), τ (1) + 1] = τ (1) − 1.
Also, since di is an excedance of τ and τ is an involution, we have |{k ∈ [τ (1)] : τ (k) ∈
[τ (1)]}| ≥ 2, which implies
τ [τ (1), τ (1) + 1] ≤ τ (1) − 2.
So σ [τ (1), τ (1) + 1] > τ [τ (1), τ (1) + 1], but σ B τ and, by Proposition 2.5, this is a
contradiction. Thus (1, ci) is not an f d-rise.
For the second part, set χ = σ (1, ci) and R = [1, ci − 1] × [σ (1) + 1, σ (ci)]. For every
(h, k) ∈ [n]2 , we have
σ [h, k] + 1, if (h, k) ∈ R,
χ [h, k] =
σ [h, k],
if (h, k) ∈
/ R.
Thus, to prove that χ ≤ B τ , we only have to show that τ [h, k] ≥ σ [h, k] + 1 for every
(h, k) ∈ R. But if (h, k) ∈ R, then we have
σ [h, k] = σ [h, τ (1) + 1] ≤ τ [h, τ (1) + 1] ≤ τ [h, k] − 1,
so χ ≤ B τ .
Proposition 4.3 allows us to give the following definitions.
Definition 4.4 Let σ, τ ∈ Invol(n), with σ B τ . The minimal covering rise of σ with
respect to τ , denoted by mcrτ (σ ) (or simply mcr ) is
mcrτ (σ ) = (di, ci).
The minimal covering transformation of σ with respect to τ , denoted by mctτ (σ ) (or
simply mct) is the covering transformation of σ with respect to the minimal covering rise:
mctτ (σ ) = ctmcr (σ ) = ct(di,ci) (σ ).
We can now give our main result concerning ed-rises.
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Theorem 4.5
Let σ, τ ∈ Invol(n), with σ B τ . If mcrτ (σ ) is an ed-rise then
mctτ (σ ) ≤ B τ.
Proof: We may assume, without loss of generality, that di = 1. Since (di, ci) is an ed-rise,
and by the definitions of di and ci, we have that
1 < σ (1) < σ (ci) ≤ τ (1) < ci.
Set a = σ (1), b = σ (ci), c = τ (1) and d = ci, so 1 < a < b ≤ c < d. Then
mctτ (σ ) = σ (1, d)(a, b)
Set χ = mctτ (σ ). We have σ ≤ B τ and we want to show that χ ≤ B τ . Let us consider the
following subsets of [n]2 :
R1a = [1, a − 1] × [a + 1, b], Ra1 = [a, b − 1] × [2, a],
Raa = [a, b − 1] × [a + 1, b],
Rba = [b, c − 1] × [a + 1, b], Rab = [a, b − 1] × [b + 1, c],
Rca = [c, d − 1] × [a + 1, b], Rac = [a, b − 1] × [c + 1, d],
Rhor = [1, d − 1] × [a + 1, c], Rver = [a, c − 1] × [2, d].
The situation is illustrated in figure 3, where the essential dots of the diagrams of σ , τ
and mctτ (σ ), and the sets just defined are represented. Note that, by the definition of ci,
there are no dots of the diagram of σ in Rhor and the only dot in Rver is that in the cell (b, d).
Also note that, for every (h, k) ∈ [n]2 , we have

σ [h, k] + 2, if (h, k) ∈ Raa ,



Rxa ∪
Ray ,
χ [h, k] = σ [h, k] + 1, if (h, k) ∈

x∈{1,b,c}
y∈{1,b,c}


σ [h, k],
otherwise.
Furthermore, we have
Raa ⊆ Rhor ∩ Rver ,
x∈{1,b,c}
Rxa ⊆ Rhor ,
Ray ⊆ Rver .
y∈{1,b,c}
Therefore, to prove that χ ≤ B τ , it suffices to show that τ [h, k] ≥ σ [h, k] + 2 if (h, k) ∈
Rhor ∩ Rver and that τ [h, k] ≥ σ [h, k] + 1 if (h, k) ∈ Rhor ∪ Rver .
Let (h, k) ∈ Rhor . Then
σ [h, c + 1] = σ [h, k],
τ [h, c + 1] ≤ τ [h, k] − 1.
BRUHAT ORDER OF THE INVOLUTIONS OF Sn
Figure 3.
253
Proof of Theorem 4.5.
So, since σ [h, c + 1] ≤ τ [h, c + 1], it follows that τ [h, k] ≥ σ [h, k] + 1.
Let (h, k) ∈ Rver . Then
σ [c, k] = σ [h, k] + c − h,
τ [c, k] = τ [h, k] + |{i ∈ [h + 1, c − 1] : τ (i) ≥ k}|
≤ τ [h, k] + c − h − 1.
So, since σ [c, k] ≤ τ [c, k], it again follows that τ [h, k] ≥ σ [h, k] + 1.
Finally, let (h, k) ∈ Rhor ∩ Rver . Then
σ [c, c + 1] = σ [h, k] + c − h,
τ [c, c + 1] ≤ τ [c, k] − 1 ≤ τ [h, k] + c − h − 2.
So, since σ [c, c + 1] ≤ τ [c, c + 1], we have τ [h, k] ≥ σ [h, k] + 2.
The preceding result has the following important consequence.
Corollary 4.6 Let σ, τ ∈ Invol(n), with σ B τ . Then
mctτ (σ ) ≤ B τ.
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Proof: If mcr is an ed-rise, the result follows from Theorem 4.5. In any case, by Proposition 4.3, (di, ci) is a suitable rise of σ such that σ (di, ci) ≤ B τ . So, if mcr is not an ed-rise,
the result follows from Lemma 3.3.
5.
Invol(n) is graded
In this section we prove the first main result of this work, namely that Invol(n) is a graded
poset, and we determine explicitly its rank function. In order to do this we first give a
characterization of the covering relation in Invol(n), in terms of suitable rises and covering
transformation.
Theorem 5.1 Let σ, τ ∈ Invol(n). Then τ covers σ in Invol(n) if and only if τ = ct(i, j) (σ ),
for some suitable rise (i, j) of σ .
Proof: Let τ = ct(i, j) (σ ), for some suitable rise (i, j) of σ . Let us examine all the
six possible cases (see Table 1). In case 1, τ = σ (i, j) covers σ in Sn so in particular
in Invol(n). In case 2, τ = σ (i, j)( j, σ ( j)); in case 3, τ = σ (i, j)( j, σ (i)); in cases 4
and 6, τ = σ (i, j)(σ (i), σ ( j)). In all these four cases, τ is greater than σ in the Bruhat
order, and has distance 2 from it in the Hasse diagram of Sn . The permutations in the interval
[σ, τ ] of Sn covering σ cannot be involutions, so τ covers σ in Invol(n). Finally, in case 5,
τ = σ (i, j)( j, σ ( j))(σ ( j), σ (i)) is greater than σ in the Bruhat order and has distance 3
from it in the Hasse diagram of Sn . In this case the permutations in the interval [σ, τ ] of Sn
covering σ , and those covered by τ cannot be involutions, so again τ covers σ in Invol(n).
On the other hand, if τ = ct(i, j) (σ ) for every suitable rise (i, j) of σ , then, by Corollary 4.6,
we have
σ B mctτ (σ ) B τ.
Thus τ does not cover σ in Invol(n).
As an example of application of Theorem 5.1, consider σ = 321654 ∈ Invol(6). The
suitable rises of σ are (1, 4), (1, 5), (1, 6), (2, 4) and (2, 5), and we have ct(1,4) (σ ) = 623451,
ct(1,5) (σ ) = 523614, ct(1,6) (σ ) = 426153, ct(2,4) (σ ) = 361452 and ct(2,5) (σ ) = 351624. So
{τ ∈ Invol(n) : σ τ in Invol(n)} = {623451, 523614, 426153, 361452, 351624}.
We can now state and prove the main result of this section.
Theorem 5.2 The poset Invol(n) is graded, with rank function ρ given by
ρ(σ ) =
inv(σ ) + exc(σ )
,
2
BRUHAT ORDER OF THE INVOLUTIONS OF Sn
255
for every σ ∈ Invol(n). In particular Invol(n) has rank
ρ(Invol(n)) =
n2
.
4
Proof: By Theorem 5.5, it suffices to show that
ρ ct(i, j) (σ ) = ρ (σ ) + 1,
for every σ ∈ Invol(n) and for every suitable rise (i, j) of σ .
As we have already seen in the proof of Theorem 5.1, the increase in the number of
inversions going from σ to ct(i, j) (σ ) is 1 in case 1, it is 2 in cases 2, 3, 4, 6, and it is 3 in
case 5. On the other hand in case 1 the involution ct(i, j) (σ ) has one excedance more than σ ,
in cases 2, 3, 4, 6 the number of excedances does not change, and in case 5 the involution
ct(i, j) (σ ) has one excedance less than σ . Therefore, in each case ρ increases by 1.
For the second part, it suffices to observe that the maximum of Sn , which is also the
maximum of Invol(n), has n/2 excedances and n(n − 1)/2 inversions.
6.
Invol(n) is EL-shellable
In this section we prove that the poset Invol(n) is E L-shellable, defining a particular edgelabelling of Invol(n), which we call “standard”, and showing that it is an E L-labelling.
Theorem 5.1 allows us to give the following definition.
Definition 6.1 The standard edge-labelling of Invol(n), with values in the set {(i, j) ∈
[n]2 : i < j} (totally ordered by the lexicographic order), is defined in the following way:
for every σ, τ ∈ Invol(n) such that τ covers σ in Invol(n), if (i, j) is the suitable rise of σ
such that τ = ct(i, j) (σ ), then we set
λ(σ, τ ) = (i, j).
We can now prove the E L-shellability of Invol(n).
Theorem 6.2 The poset Invol(n) is E L-shellable, having the standard edge-labelling as
an E L-labelling.
Proof: Suppose we label the edges of the Hasse diagram of Invol(n) with the standard
labelling.
Let σ, τ ∈ Invol(n), with σ B τ . Consider the saturated chain from σ to τ
σ = σ0 σ1 · · · σk = τ,
256
INCITTI
defined by
σi = mctτ (σi−1 ),
for every i ∈ [k]. Corollary 4.6 ensures that σi ≤ B τ for every i ∈ [k − 1].
By the definitions of mct, di and ci, this chain has, among all the saturated chains from
σ to τ , the minimal labelling in the lexicographic order.
We now prove that it has increasing labels. Suppose, by contradiction, that at a certain
step there is a decrease in the labels. We may assume, without loss of generality, that this
happens at the first step. Thus
σ1 = mctτ (σ ) = ct(di,ci) (σ )
and
σ2 = ct(i, j) (σ1 ),
with (i, j) < L (di, ci). So either i < di or i = di and j < ci. If i < di, since σ and τ
must differ at the index i, the minimality of di is contradicted. If i = di and j < ci, since
σ ( j) ∈ [σ (di) + 1, τ (di)], the minimality of ci is contradicted.
It remains to show that any other saturated chain from σ to τ , different from the previuos
one, has at least one decrease. Let
σ = τ0 τ1 · · · τk = τ
be such a saturated chain. Set h = min{i ∈ [k − 1] : τi = σi }, di = di τ (σh−1 ) and
ci = ci τ (σh−1 ). So
σh = mctτ (σh−1 ) = ct(di,ci) (σh−1 )
and
τh = ct(i, j) (σh−1 ) ,
for some suitable rise (i, j) of σh−1 different from (di, ci) and lexicographically greater
than it. So either di < i or di = i and ci < j.
If di < i, then in the covering relations τh τh+1 · · · τk = τ there must be at least
one with label containing di, so lower than (i, j).
Suppose di = i and ci < j. Since the dot in column di has to move from row σ (di) to
row τ (di) and because of the presence in the diagram of σ of the dot in the cell (ci, σ (ci)),
in the covering relations τh τh+1 · · · τk = τ either there is one with label (di, ci),
so lower than (i, j), or there is one with label starting with ci, followed by one with label
starting with di, so again with a decrease.
As a consequence, by Theorem 2.1, we have the following.
Corollary 6.3
The poset Invol(n) is Cohen-Macaulay.
BRUHAT ORDER OF THE INVOLUTIONS OF Sn
7.
257
Invol(n) is Eulerian
In this section we prove that the poset Invol(n) is Eulerian. In order to do this, we introduce
some notions which, in some sense, invert those introduced in Sections 3 and 4.
Definition 7.1 Let τ ∈ Invol(n). An inversion (i, j) of τ is inv-suitable if (i, j) is a suitable
rise of some σ ∈ Invol(n) and ct(i, j) (σ ) = τ . We call such a σ (obviously unique) the inverse
covering transformation of τ with respect to (i, j) and we denote it by ict(i, j) (τ ).
Obviously ict(i, j) (ct(i, j) (σ )) = σ and ct(i, j) (ict(i, j) (τ )) = τ . We summarize the action
of the inverse covering transformation on the diagram of an involution in Table 2, with a
notation similar to that used in Table 1. In this case the black dots denote τ and the white
dots denote ict(i, j) (τ ).
Definition 7.2 Let σ, τ ∈ Invol(n), with σ B τ . The minimal covering inversion of τ
with respect to σ , denoted by mci σ (τ ) (or simply mci), is the minimal (in the lexicographic
order) inv-suitable inversion (i, j) of τ such that σ ≤ B ict(i, j) (τ ).
Table 2.
Inverse covering transformation.
Case
1
2
3
4
5
6
Move
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The minimal inverse covering tranformation of τ with respect to σ , denoted by mictσ (τ )
(or simply mict), is the inverse covering transformation of τ with respect to the minimal
covering inversion:
mictσ (τ ) = ictmci (τ ).
We can now prove that the condition of Theorem 2.2 holds for the poset Invol(n), and
thus that it is Eulerian.
Theorem 7.3
The poset Invol(n) is Eulerian.
Proof: Suppose we label the edges of the Hasse diagram of Invol(n) with the standard
labelling.
Let σ, τ ∈ Invol(n), with σ B τ . By Theorem 2.2, we only have to show that there is
exactly one saturated chain from σ to τ with decreasing labels.
In this proof we use the following terminology: if (i, j) is an inv-suitable inversion of τ
we call it simply a move for τ , precisely an a-move, with a ∈ [6], if we are in case a of
Table 2. Furthermore, if ict(i, j) (τ ) = σ , then we write
τ σ.
(i, j)
We divide the proof in two parts. We first prove that there is at least one saturated chain
from σ to τ with decreasing labels. Consider the descending chain
τ = σ0 σ1 · · · σk = σ,
defined by
σi = mictσ (σi−1 ),
for every i ∈ [k]. We claim that it has increasing labels (so the corresponding ascending
chain will have decreasing labels). Suppose, by contradiction, that at a certain step there is
a decrease in the labels. We may assume, without loss of generality, that this happens at the
first step. So
σ0 σ1 σ2 ,
(i, j)
(i , j )
with (i , j ) < L (i, j). There are two cases: either i < i or i = i and j < j.
If i < i, then (i, j) cannot be the minimal choice for σ0 , since σ0 must have an inv-suitable
inversion containing i . This contradicts the definition of the chain.
If i = i and j < j, again (i, j) cannot be the minimal choice for σ0 . The proof of this
fact is a case-by-case verification, depending on the type of (i, j). We show some cases,
leaving the others to the reader.
BRUHAT ORDER OF THE INVOLUTIONS OF Sn
Figure 4.
259
Proof of Theorem 7.3.
First of all note that (i, j) cannot be a 1-move or a 2-move, in fact in this case we could
not apply to σ1 a move (i, j ), with j < j. If (i, j) is a 3-move and (i, j ) is a 1-move, the
situation is illustrated in figure 4(a): if we apply to σ0 the two moves (i, j ) and ( j , j) (in
this order), we again reach σ2 :
σ0 σ1 σ2 .
(i, j )
( j , j)
But (i, j ) < L (i, j), so (i, j) is not the minimal choice for σ0 . In the picture we represent
a pair of moves by colouring the areas “enclosed” in the moves, with a lighter grey for
the first move and a darker grey for the second one; the arrow represents the possibility of
substituting a pair of moves with another pair reaching the same involution. Figure 4(b) is
the synthetic version of figure 4(a). If (i, j) is a 3-move, all other cases are synthetically
described in figure 4(c–f), with the notation described above. If (i, j) is a 4, 5 or 6-move
the reasoning is similar. In each case we get a contradiction.
260
INCITTI
We now prove that any other saturated chain from σ to τ , different from the previous
one, has at least one increase. Let
τ = τ 0 τ1 · · · τ k = σ
be a saturated descending chain from τ to σ , different from the previous one. We will prove
that in it there is at least one decrease (so the corresponding ascending chain will have at
least one increase). Set h = min{i ∈ [k − 1] : τi = σi } and mci σ (σh−1 ) = (i, j). We have
σh = ict(i, j) (σh−1 )
and
τh = ict(i , j ) (σh−1 ),
for some inv-suitable inversion (i , j ) of σh−1 different from (i, j) and lexicographically
greater than it. So either i < i or i = i and j < j .
If i < i , then in the covering relations τh τh+1 · · · τk = σ there must be one with
label containing i, so lower than (i , j ).
Suppose i = i and j < j . We want to show that in the labelling of the descending chain
σh−1 τh τh+1 · · · τk = σ there is at least one decrease. Suppose, by contradiction,
that it has increasing labels. If l = min{s ∈ [h, k] : τs (i) = σ (i)}, then
σh−1 τh τh+1 · · · τl−1 τl ,
(i, j )
(i, j1 )
(i, j2 )
(i, jl−h )
with ( j <) j < j1 < j2 < · · · < jl−h . We want to show that this is in contradiction with
the fact that (i, j) is an inv-suitable inversion of σh−1 . The proof of this fact is again a
case-by-case verification, depending on the type of (i, j).
If (i, j) is a 1-move, a 3-move or a 6-move, then the (i, jr )’s can only be 6-moves, and
none of these can send the dot in column i on or below row σh (i). But σ (i) ≤ σh (i), and we
have a contradiction.
If (i, j) is a 2-move, a 4-move or a 5-move, then the sequence of the (i, jr )’s can only be
realized by a sequence (possibly empty) of 4-moves, possibly followed by a 3 or a 5-move
(but not both) and then by a sequence (possibly empty) of 6-moves. If (i, j) is a 4-move,
then none of these moves can send the dot in column i on or below row σh (i). If (i, j) is a
2-move or a 5-move, then the 3 or 5-move is the only one that can move the dot in column
i on row j, and in this case none of the following 6-moves can move that dot on or below
row σh (i). But, as before, σ (i) ≤ σh (i), so in each case we get a contradiction.
Furthermore, by Theorem 2.3, we can conclude the following.
Corollary 7.4 The poset Invol(n) is Gorenstein.
BRUHAT ORDER OF THE INVOLUTIONS OF Sn
261
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